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What percent of a drug is contained in a mixture of powder consisting of 0.5 kg, containing 0.038% of a drug, and 10 kg, containing 0.043% of a drug?

Respuesta :

Answer:

[tex]0.0427~%[/tex]

Explanation:

In this question, we have to start with the calculation of the amount of drug in each powder:

Powder A: Total mass of 0.5 Kg percentage of 0.038%

[tex]0.5~Kg~of~powder\frac{0.038}{100}=0.00019~Kg~of~drug[/tex]

Powder B: Total mass of 10 Kg percentage of 0.043%

[tex]10~Kg~of~powder\frac{0.043}{100}=0.0043~Kg~of~drug[/tex]

The total mass of powder would be:

[tex]10+0.5=10.5~Kg[/tex]

The total mass of drug would be:

[tex]0.0043+0.00019=0.00449~Kg[/tex]

Now we can calculate the percentage:

[tex]\frac{0.0043Kg}{10.5Kg}*100=0.0427%[/tex]

I hope it helps!